Finite Abelian Groups

Most of the time, "classify all groups of order $n$" is hopeless. For Abelian groups it isn't: there is a complete list, and you can write it down by factoring $n$ and counting. This page states the theorem that makes that possible, shows how to use it, and then proves the finite version. On this page $\mathbb{Z}_n$ means the integers mod $n$ under addition, so $\mathbb{Z}_n \cong C_n$.

Free Abelian Groups and Invariant Factors

The building blocks are cyclic groups: finite ones $\mathbb{Z}_n$, and the infinite one $\mathbb{Z}$. Copies of $\mathbb{Z}$ get a name of their own.

($\mathbb{Z}^0$ is the trivial group.) The Fundamental Theorem says that every finitely generated Abelian group (F.G.A.G.) is built from these blocks in exactly one tidy way.

Condition (ii) is what makes the answer unique. Without it, $\mathbb{Z}_2 \times \mathbb{Z}_3$ and $\mathbb{Z}_6$ would be two different-looking names for the same group. With it, only $\mathbb{Z}_6$ is allowed, because $2 \nmid 3$. The numbers in the theorem have names.

There's also a compact way to quote the whole list at once.

For example, $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}_2 \times \mathbb{Z}_6$ has free rank $2$ and is of type $(2, 6)$. A finite group has free rank $0$, since even one copy of $\mathbb{Z}$ makes the group infinite. From here on, we only prove the finite case. The $\mathbb{Z}^r$ part needs different tools and is not proved on this page.

Elementary Divisors

The invariant factors stack everything into as few cyclic groups as possible. The other extreme breaks everything down into cyclic groups of prime-power order. This is the finite-groups version, ignoring free rank $r$.

Part (i) splits $G$ up one prime at a time, and part (ii) splits each prime's piece into cyclic groups. Notice that $\beta_{i1} + \beta_{i2} + \dots + \beta_{it_i} = \alpha_i$, since the orders have to multiply to $|G_i| = p_i^{\alpha_i}$. We'll see that both versions of the theorem are equivalent: each list can be computed from the other.

Counting Abelian Groups

The theorem turns "list every Abelian group of order $n$" into arithmetic. Start with the easiest orders.

The choices come from repeated primes. For a single prime, the only freedom is how to split the exponent.

Writing a number as a sum, ignoring order, is a classic idea in number theory.

So $P(1) = 1$, $P(2) = 2$, $P(3) = 3$, $P(4) = 5$, and Example 2 says:

• 2d) $|G| = p^k$. The number of different Abelian groups of order $p^k$ is $P(k)$, the number of ways of writing $k$ as a sum.

$P(k)$ grows fast. Hardy and Ramanujan found that $$P(k) \sim \frac{1}{4k\sqrt{3}} \exp\left(\pi\sqrt{\frac{2k}{3}}\right) \quad \text{as } k \rightarrow \infty.$$

With several primes, part (i) of the theorem lets each prime choose its partition independently.

• 3) $|G| = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$, $p_1 < \dots < p_k$ primes, $\alpha_i \in \mathbb{N}$. By the elementary divisor decomposition, $G \cong G_1 \times \dots \times G_k$, $|G_i| = p_i^{\alpha_i}$, and there are $P(\alpha_i)$ choices for each $G_i$. So the number of Abelian groups of order $n$ is $$\prod_{i=1}^{k} P(\alpha_i).$$

Below, each slider sets the exponent of one prime, and each prime's panel shows its partitions as stacks of blocks: a column of $\beta$ blocks is one factor $\mathbb{Z}_{p^\beta}$. Pick one stack per prime to build a group.

Notice that the group is cyclic exactly when every prime picks its one-column stack. That's Example 1 again, now for any $n$: $\mathbb{Z}_n$ is always on the list, and it's the only cyclic one.

Switching Between the Two Forms

Everything rests on one fact about cyclic groups.

Read left to right, it cracks a cyclic group into prime-power pieces. Read right to left, it glues pieces for different primes back together. The first example goes from invariant factors to elementary divisors.

Going the other way, glue together the largest power of each prime, then the next largest, and so on.

The general case works the same way. Write the elementary divisors in one row per prime, each row in increasing order and lined up on the right. Multiply down each column to get one invariant factor per column. Reading the columns left to right, every prime power in a column is at most the one to its right, so each invariant factor divides the next one. Going backwards, factoring each invariant factor into prime powers recovers the rows. This is why the two versions of the FTFGAG are equivalent.

Watch it happen. The starting group doesn't even have to be in either form.

The $\mathbb{Z}_{10} \times \mathbb{Z}_{15} \times \mathbb{Z}_6$ preset is worth a second look: three factors went in and two came out, so a different-looking product of cyclic groups can be the same group.

Cauchy's Theorem

Now for the proof of the finite version. We'll prove the elementary divisor decomposition, since the two versions are equivalent. First, a theorem we will use in the proof, and one that is worth knowing in its own right.

Notice that $G$ doesn't have to be Abelian. The proof, due to James McKay, counts one set in two ways.

Show proof

Proof. Let $S = \{(g_1, \dots, g_p) : g_1, \dots, g_p \in G,\ g_1 \cdots g_p = e\}$. Define $\sim$ on $S$ as follows: $x \sim y \iff x$ and $y$ are cyclic permutations of each other, i.e. $$(g_1, \dots, g_p) \sim (g_2, \dots, g_p, g_1) \sim \dots$$ Rotating stays inside $S$: if $g_1 g_2 \cdots g_p = e$ then $g_2 \cdots g_p = g_1^{-1}$, so $g_2 \cdots g_p g_1 = e$. This is an equivalence relation. Calculate $|S|$ in two ways.

Way 1: by equivalence classes. There are $2$ types of equivalence classes: classes with $1$ element (e.g. $(e, \dots, e)$), and classes with $p$ different elements. To see why, suppose rotating $x = (g_1, \dots, g_p)$ by $j$ places gives $x$ back, for some $0 < j < p$. Then $g_i = g_{i+j}$ for every $i$ (indices mod $p$). Since $p$ is prime, $(j, p) = 1$, so stepping by $j$ visits every index, and all the $g_i$ are equal. So either $x$ is constant, or its $p$ rotations are all different (if rotations by $a$ and $b$ agreed, rotation by $b - a$ would fix $x$). So $$|S| = 1 \cdot (\#\text{ of equiv. classes w/ } 1 \text{ elem.}) + p \cdot (\#\text{ of equiv. classes w/ } p \text{ elems.}) = k + pl.$$ The one-element classes are the constant tuples $(g, \dots, g)$ with $g^p = e$. Note $k \ge 1$, because of $(e, \dots, e)$.

Way 2: directly. $g_1, \dots, g_{p-1}$ can be anything, and then $g_p$ is forced: $$S = \{(g_1, \dots, g_{p-1}, (g_1 \cdots g_{p-1})^{-1}) : g_1, \dots, g_{p-1} \in G\} \implies |S| = |G|^{p-1} = p^{p-1} \cdot m$$ for some $m \in \mathbb{N}$ (since $p \mid |G|$).

Compare the formulas: $$k + pl = p^{p-1} m \implies p \mid k \implies k > 1 \implies \exists\, g \in G \setminus \{e\} \text{ s.t. } g^p = e \implies |g| = p,$$ since $|g|$ divides $p$ and isn't $1$. $\blacksquare$

The necklaces below are the elements of $S$: read clockwise from the top, the beads are $g_1, g_2, \dots, g_p$. Turning a necklace is a cyclic permutation, so an equivalence class is every different-looking necklace you can get by turning one.

The proof only ever uses one fact about the one-element classes: there are at least $p$ of them. In $S_3$ with $p = 3$, they are $(e, e, e)$, $((1\ 2\ 3), (1\ 2\ 3), (1\ 2\ 3))$, and $((1\ 3\ 2), (1\ 3\ 2), (1\ 3\ 2))$, which hands us the two elements of order $3$.

Proof of Part (i): Splitting by Primes

Recall the setup: $|G| = p_1^{\alpha_1} \cdots p_k^{\alpha_k}$, $p_1 < \dots < p_k$ primes, $\alpha_i \in \mathbb{N}$, and we want to show that $G \cong G_1 \times \dots \times G_k$ with $|G_i| = p_i^{\alpha_i}$. The plan is to show that for every $1 \le i \le k$, $G$ has a unique subgroup $G_i$ of order $p_i^{\alpha_i}$, and then glue them together. There is only one sensible candidate:

$$G_i = \{g \in G : |g| \text{ divides } p_i^{\alpha_i}\}.$$

Show proof

• $G_i$ is a group. $e \in G_i$. Suppose $g, h \in G_i$. Then $|h^{-1}| = |h|$ divides $p_i^{\alpha_i}$. Also $|gh|$ divides $|g||h|$ (as $G$ is Abelian, $(gh)^{|g||h|} = (g^{|g|})^{|h|}(h^{|h|})^{|g|} = e$), which is a power of $p_i$, and $|gh|$ divides $|G|$ by Lagrange. So $|gh|$ is a power of $p_i$ that is at most $p_i^{\alpha_i}$, and $gh \in G_i$.
• $|G_i| = p_i^{a_i}$ for some $0 \le a_i \le \alpha_i$. By Lagrange's theorem, $|G_i|$ divides $|G|$, so $|G_i| = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ with $0 \le a_j \le \alpha_j$. If $j \neq i$ but $a_j > 0$, then $p_j \mid |G_i|$, so $G_i$ has an element of order $p_j$ by Cauchy's theorem. But $p_j$ doesn't divide $p_i^{\alpha_i}$, which contradicts the definition of $G_i$.
• We can't have $|G_i| = p_i^{a_i}$ with $a_i < \alpha_i$. If it were, consider $\overline{G_i} = G/G_i$ (a group, since every subgroup of an Abelian group is normal). Then $|\overline{G_i}| = |G|/|G_i|$, so $p_i \mid |\overline{G_i}|$, and $\overline{G_i}$ has an element of order $p_i$ by Cauchy's theorem. Call it $gG_i$. Write $|g| = p_i^c q$, where $(p_i, q) = 1$. Then $$|g^q| = \frac{|g|}{\gcd(|g|, q)} = p_i^c,$$ and $p_i^c$ divides $|G|$, so $c \le \alpha_i$ and $g^q \in G_i$ by definition. That means $(gG_i)^q = G_i$, so $|gG_i| = p_i$ divides $q$, a contradiction. Conclusion: $|G_i| = p_i^{\alpha_i}$.
• Uniqueness follows easily from the definition of $G_i$: if $H \le G$ has $|H| = p_i^{\alpha_i}$, then every $h \in H$ has $|h|$ dividing $p_i^{\alpha_i}$ (Lagrange in $H$), so $H \subseteq G_i$. The two have the same size, so $H = G_i$. $\blacksquare$

Now glue. For $i \neq j$, $G_i \cap G_j$ is a subgroup of both $G_i$ and $G_j$, so its order divides $p_i^{\alpha_i}$ and $p_j^{\alpha_j}$, and $G_i \cap G_j = \{e\}$. It's tempting to stop there and say the product of all the $G_i$ has $|G_1| \cdots |G_k|$ elements.

Finishing part (i). Let $P_j = G_1 G_2 \cdots G_j$. We show by induction on $j$ that $P_j \cong G_1 \times \dots \times G_j$ and $|P_j| = p_1^{\alpha_1} \cdots p_j^{\alpha_j}$. For $j = 1$ there's nothing to do. For the step, $P_j \cap G_{j+1}$ is a subgroup of both, so its order divides $p_1^{\alpha_1} \cdots p_j^{\alpha_j}$ and $p_{j+1}^{\alpha_{j+1}}$, which are coprime. So $P_j \cap G_{j+1} = \{e\}$. Every subgroup of $G$ is normal, so by the Internal Direct Product Theorem and the Product Formula, $$P_{j+1} = P_j G_{j+1} \cong P_j \times G_{j+1} \cong G_1 \times \dots \times G_{j+1}, \qquad |P_{j+1}| = |P_j| \cdot |G_{j+1}|.$$ At $j = k$, $|G_1 \cdots G_k| = |G_1| \cdots |G_k| = |G|$, so $G_1 \cdots G_k = G$ and $G \cong G_1 \times \dots \times G_k$. $\blacksquare$

$p$-Groups, Exponents, Elementary Abelian Groups

Each $G_i$ only contains elements whose orders are powers of $p_i$. Groups like that come up constantly, so they get a name, along with a few more definitions.

So every $G_i$ is a $p_i$-group. A finite Abelian $p$-group always has order $p^n$ for some $n$: if another prime $q$ divided its order, Cauchy's theorem would give an element of order $q$, and $q$ is not a power of $p$.

The basic examples are $\mathbb{Z}_p, \mathbb{Z}_p \times \mathbb{Z}_p, \dots$: every non-identity element has order exactly $p$.

For a finite Abelian group the exponent is easy to read off once you have the invariant factors.

Proof of Part (ii): Splitting Off a Cyclic Factor

Let $G$ be a finite Abelian $p$-group. The idea is to peel off one cyclic factor at a time: take an element $g$ of the largest possible order, and find a subgroup $K$ that fills in the rest of $G$ without overlapping $\langle g \rangle$. Such a $K$ is called a complement of $\langle g \rangle$.

Try it first. The groups below are written additively, so $e = (0, 0)$, and $\langle g \rangle$ is shaded blue. "Maximal order" turns out to matter: watch the last mission.

In the first three missions $g$ had maximal order, and a complement was always there. The Lemma says that's no accident.

The proof is by induction on the order of $G$. The induction runs through the quotient $G/H$ for a well-chosen subgroup $H$, and the hard part is choosing $H$; that is the Claim inside the proof.

Show proof

Proof of the Lemma. Write $|G| = p^n$. Proof by induction on $n$. (If $n = 0$, $G = \{e\}$ and $K = \{e\}$ works.)

Base case ($n = 1$): $|G| = p$; let $g \in G \setminus \{e\}$ and $K = \{e\}$. $\checkmark$

Inductive hypothesis: suppose the Lemma is true for all groups of order at most $p^{n-1}$.

Inductive step: suppose $|G| = p^n$, and let $g \in G$ have maximal order. If $|g| = p^n$, then $G = \langle g \rangle$; take $K = \{e\}$ and we're done. So suppose $|g| = p^l$ for some $1 \le l < n$.

Claim: there is a subgroup $H \le G$, $|H| > 1$, satisfying (i) $\langle g \rangle \cap H = \{e\}$, and (ii) $|gH| = p^l$ in $G/H$.

Assuming the Claim: first note that $gH$ is an element of maximal order in $G/H$. Indeed, for any $yH \in G/H$, $(yH)^{|y|} = y^{|y|}H = H$, so $|yH| \le |y| \le p^l = |gH|$. Also $|G/H| = |G|/|H| < |G|$, and $G/H$ is again an Abelian $p$-group. So by the inductive hypothesis, $\exists\, \overline{K} \le G/H$ such that $G/H \cong \langle gH \rangle \times \overline{K}$ and $\langle gH \rangle \cap \overline{K} = \{eH\}$. By the 4th Isomorphism Theorem, $\exists\, K \le G$ with $H \le K$ such that $\overline{K} = K/H$.

Subclaim: $\langle g \rangle \cap K = \{e\}$.

Proof. Suppose $x \in \langle g \rangle \cap K$. Then $xH \in \langle gH \rangle \cap K/H = \langle gH \rangle \cap \overline{K} = \{eH\}$, so $x \in H$. So $x \in \langle g \rangle \cap H = \{e\}$, and $x = e$. $\checkmark$

Now count. $G/H \cong \langle gH \rangle \times K/H$, so $$|G/H| = |\langle gH \rangle| \cdot |K/H| \implies \frac{|G|}{|H|} = |g| \cdot \frac{|K|}{|H|} \implies |G| = |\langle g \rangle| \, |K|.$$ By the Product Formula and the Subclaim, $|\langle g \rangle K| = |\langle g \rangle| |K| = |G|$, so $\langle g \rangle K = G$. Both subgroups are normal ($G$ is Abelian) and they meet in $\{e\}$, so by the Internal Direct Product Theorem, $G = \langle g \rangle K \cong \langle g \rangle \times K$. So this proves the Lemma, assuming the Claim. $\checkmark$

---

Proof of the Claim. Since $|g| = p^l < p^n$, $G \neq \langle g \rangle$. Let $h \in G \setminus \langle g \rangle$ be an element of smallest possible order. Note $|h| = p^i$ for some $1 \le i \le l$ ($h \neq e$, and $g$ has maximal order). Then $$|h^p| = \frac{|h|}{\gcd(|h|, p)} = \frac{|h|}{p} < |h| \implies h^p \in \langle g \rangle,$$ by the choice of $h$. So $h^p = g^a$ for some $a \in \mathbb{N}$. Next, $|h^p| = |h|/p \le p^l/p < |g|$, and also $$|h^p| = |g^a| = \frac{|g|}{\gcd(|g|, a)}.$$ So $\gcd(p^l, a) > 1$, which means $p \mid a$, and $h^p = g^{bp}$ for some $b \in \mathbb{N}$. Let $x = g^{-b} h \in G$. Then:

1. $x^p = g^{-bp} h^p = e$ ($G$ is Abelian).
2. $x \notin \langle g \rangle$: if $x = g^c$ for some $c$, then $h = g^{b + c} \in \langle g \rangle$, a contradiction.

In particular $x \neq e$, so $|x| = p$. Take $H = \langle x \rangle$. Then $|H| = p > 1$, and:

1. $\langle g \rangle \cap H = \{e\}$. This is a subgroup of $H \cong C_p$, so it is $\{e\}$ or $H$, and it isn't $H$ because $x \notin \langle g \rangle$.
2. $|gH| = p^l$: $(gH)^m = eH \iff g^m \in H \implies g^m \in \langle g \rangle \cap H = \{e\} \implies m = p^l k$ for some $k$. And $(gH)^{p^l} = eH$. $\checkmark$ $\blacksquare$

Mission 4 shows where the proof needs maximal order. There, $g = (2, 0)$ and $H = \langle (0, 1) \rangle$ still satisfy the Claim, but $G/H \cong \mathbb{Z}_4$, and $gH$ has order $2$ while $(1, 0) + H$ has order $4$. So $gH$ is not of maximal order in $G/H$, and the inductive hypothesis has nothing to say about it.

Finishing part (ii). Each $G_i$ from part (i) is a finite Abelian $p_i$-group of order $p_i^{\alpha_i}$. Pick $g_1 \in G_i$ of maximal order $p_i^{\beta_{i1}}$. The Lemma gives $G_i \cong \langle g_1 \rangle \times K_1 \cong \mathbb{Z}_{p_i^{\beta_{i1}}} \times K_1$, where $K_1$ is a smaller Abelian $p_i$-group. Now pick $g_2 \in K_1$ of maximal order $p_i^{\beta_{i2}}$ and apply the Lemma to $K_1$, and so on. Each $K_j$ is a subgroup of $G_i$, so its largest order is at most the previous one, giving $\beta_{i1} \ge \beta_{i2} \ge \cdots$. The orders shrink every time, so the process stops (at $K_{t_i} = \{e\}$) with $$G_i \cong \mathbb{Z}_{p_i^{\beta_{i1}}} \times \mathbb{Z}_{p_i^{\beta_{i2}}} \times \dots \times \mathbb{Z}_{p_i^{\beta_{it_i}}}. \qquad \blacksquare$$

Part (ii) also settles elementary Abelian groups completely.

Uniqueness (Part iii)

Uniqueness of the decomposition $G \cong G_1 \times \dots \times G_k$ in part (i) follows from the construction we gave: $G_i$ had to be the unique subgroup of order $p_i^{\alpha_i}$, namely $\{g \in G : |g| \text{ divides } p_i^{\alpha_i}\}$, and that set is determined by $G$ alone.

Uniqueness of the decomposition in part (ii) follows from comparing elements of given orders. For example, $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^2}$ and $\mathbb{Z}_{p^3} \times \mathbb{Z}_p$ both have order $p^4$, but only the second has an element of order $p^3$. In general, counting the elements with $g^{p^j} = e$ for each $j$ recovers every $\beta_{ij}$. Making that precise is the Spicy Bonus.

Spicy Bonus

Part (iii) of the theorem was only sketched. Here it is for you to prove.